UGC NET Computer Science Computer Architecture Previous Year Questions (PYQs) – Page 1 of 2

Using Amdahl’s law:

Fraction improved = $0.30$

Speedup of improved part = $2$

Overall speedup:

$S = \frac{1}{(1 - 0.30) + \frac{0.30}{2}}$

$S = \frac{1}{0.70 + 0.15}$

$S = \frac{1}{0.85}$

$S = 1.176 \approx 1.18$

Content addressable memory is also called associative memory.

In this memory, data is searched by content instead of address.

Register $A$ contains the bit pattern and register $K$ decides which bits are to be compared.

Since this memory supports parallel searching by data association, it is called content addressable memory.

Memory capacity is

$16K \times 16$

Here, $16K$ means number of memory locations.

$16K=16 \times 1024$

$=2^4 \times 2^{10}$

$=2^{14}$

So, number of address lines required is

$14$

Each memory location stores $16$ bits.

So, number of data lines required is

$16$

Therefore, address lines and data lines are

$14,\ 16$

Total execution time is

$100$ seconds

Time taken by multiplication operation is

$80$ seconds

Time taken by other operations is

$100-80=20$ seconds

Now, the program is required to run $4$ times faster.

So, new execution time should be

$\dfrac{100}{4}=25$ seconds

Other operations still take

$20$ seconds

So, time left for multiplication operation is

$25-20=5$ seconds

Originally multiplication operation takes

$80$ seconds

Now it should take

$5$ seconds

So, required improvement in multiplication speed is

$\dfrac{80}{5}=16$

Hence, multiplication operation must be improved by $16$ times.

Resolution of raster system is

$640 \times 480$

Total number of pixels is

$640 \times 480=307200$

Each pixel requires $12$ bits.

So, total memory required is

$307200 \times 12=3686400$ bits

Convert bits into bytes:

$\dfrac{3686400}{8}=460800$ bytes

Now,

$1\text{ KB}=1024\text{ bytes}$

So,

$\dfrac{460800}{1024}=450\text{ KB}$

Given matrix is

When this matrix is applied on point $(x,y)$, then the point becomes $(x,-y)$.

This means the sign of $x$ remains same and the sign of $y$ changes.

So, this represents reflection about the $X$-axis.

Therefore, statement $S1$ is true.

But reflection about the $X$-axis is not a rotation matrix.

Therefore, statement $S2$ is false.

Hence, the correct answer is Option 2.

Orthographic transformation preserves parallelism.

So, parallel lines remain parallel after orthographic projection.

Therefore, statement $P$ is true.

Orthographic transformation is also an affine transformation because it preserves points, lines and parallelism.

Therefore, statement $Q$ is also true.

In the Phong reflectance model, specular reflection depends on how close the viewer direction is to the reflected light direction.

If the reflected vector and the view vector are close to each other, the specular highlight becomes stronger.

So, the strength of the specular highlight is determined by the angle between the reflected vector and the view vector.